Question: The equation of a circle $C$ is $x^2+y^2-2x+14y+34 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-2x) + (y^2+14y) = -34$ $(x^2-2x+1) + (y^2+14y+49) = -34 + 1 + 49$ $(x-1)^{2} + (y+7)^{2} = 16 = 4^2$ Thus, $(h, k) = (1, -7)$ and $r = 4$.